Challenging if $\lim_{n\to\infty}\sum_{k=0}^n(1-q)q^k$ is equal to 1 If I suppose $U(0<q<1)$ as $$U(q) = \lim_{n\to\infty}\sum_{k=0}^n(1-q)q^k$$
Do $$A = \lim_{q\to0}U(q)$$ and $$B = \lim_{q\to1}U(q)$$ exist? if so, what we can say about them? e.g. are they really reach each other at value $1$ in $\infty$?
EDIT: SIMPLIFIED
Suppose we both are at value 0, then you start to go to value 1 with speed $q\to0$ and I also but with speed $q\to1$. Do we both reach each other at 1 in $\infty$ while I always am at front of you? it's hard for me to understand this!
 A: Before taking the limit of the function U(q) with respect to variable "q", firstly you should find the value of the function. Then you can take the limit. So
$$U(q)=\lim\limits_{n→∞}\sum_{k=0}^n(1−q)q^k=(1−q)\lim\limits_{n→∞}\sum_{0}^nq^k=\frac{1-q}{1-q}=1$$
Then your function is a constant function such that $ U(q)=1$. So it is independent of limit with respect to variable $q$.
A: This is a very subtle and important issue in mathematics.  Both $A$ and $B$ are in fact equal to one.  The issue here is that it matters in which order you take the limits.  For instance,
$$1=\lim_{q\to 0} \lim_{n\to\infty} \sum_{k=0}^n (1-q) q^k \neq
 \lim_{n\to\infty} \lim_{q\to 0} \sum_{k=0}^n (1-q) q^k=0.$$
In the double limit to the right of the inequality you first take the limit of $q\to 0$, which makes all summands equal to zero and the sum of zeroes equals zero.  In the double limit on the left you first take the limit as $n\to\infty$ which makes the limit equal to one after which you take the limit with respect to $q$, but a limit of a constant sequence equals the constant.
